If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:10:24

sometimes light seems to act as a wave and sometimes light seems to act as a particle and an example of this would be the photoelectric effect as described by Einstein so let's say you had a piece of metal and we know the metal has electrons I'm going to go ahead and draw one electron in here and this electron is bound to the metal because it's attracted to the positive charges in the nucleus if you shine a light on the metal so the right kind of light with the right kind of frequency you can actually knock some of those electrons loose which causes a current of electrons to flow so this is kind of like a collision between two particles if we think about light as being a particle so I'm going to draw in a particle of light which we call a photon so this is massless and the photon is going to hit this electron and if the photon has enough energy it can free the electron right so we can knock it loose and so let me go ahead and show that so here we're showing the electron being knocked loose and so the electron is moving in let's just say this direction with with some velocity V and if the electron has mass M we know that there's a kinetic energy right the kinetic energy of the electron would be equal to one-half MV squared this freed electron is usually referred to now as a photo electron so one photon creates one photo electron so one particle hits another particle and if you think about this in terms of classical physics right you could think about energy being conserved so the energy the photon the energy that went in let me go ahead and write this here so the energy of the photon the energy that went in what happened to that energy some of the energy was needed to free the electron right so the electron was bound and some of the energy freed the electrons I'm going to call that II not the energy that freed the electron and then the rest of that energy must have gone into the kinetic energy of the electron and so we can write here kinetic energy of the photo electron that was produced so kinetic energy of the photo electron so let's say you wanted to solve for the kinetic and G of that photoelectron so that would be very simple just be kinetic energy would be equal to the energy of the photon energy the photon minus the energy that was necessary to free the electron from the metallic surface and this e naught here I'm calling it a nought you might see it you might see it written differently a different symbol but this is the work function so let me go ahead and write work function here and the work function is different for every kind of metal so the it's the minimum amount of energy that's necessary to free the electron and so obviously that's going to be different depending on what metal you're talking about all right let's do a problem now that we understand the general idea of the photoelectric effect let's look at what this problem asks us so the problem says if a photon of wavelength 525 nanometers hits metallic cesium and so here's the work function for metallic cesium what is the velocity of the photo electron produced so they want to know the velocity the photo electron produce which we know is hiding in the kinetic energy right here and we also know what the work function is so we know what enon is here what we don't know is the energy of the photon so that's what we need to calculate first and so the energy of the photon energy the photon is equal to H which is Planck's constant times the frequency which is usually symbolized by nu so we have the frequency but they gave us the wavelength in the problem here they gave us wavelength so we need to relate frequency to wavelength and that's related by C which is the speed of light is equal to lambda times nu so C is the speed of light and that's equal to the frequency times the wavelength so we can now we can substitute in for the frequency alright because we can just use this equation and say that the frequency is equal to the speed of light divided by the wavelength C is B the frequency is equal to speed of light over lambda so we can plug that into here and so now we have the energy of the photon is equal to HC over lambda and we can plug in those numbers H is Planck's constant which is six point six two six times ten to the negative 34 so times 10 to the negative 34 here C is the speed of light which is 2.99 H times 10 to the 8th meters over seconds and all over lambda lambda is the wavelength that's 525 nanometers so 525 times 10 to the negative ninth meters all right so let's get out our calculator and calculate the energy of the photon here so let's go ahead and do that math so we have six point six two six times 10 to the negative 34 and we're going to multiply that number by the speed of light two point nine nine eight times ten to the eighth and we get that number we're going to divide it by the wavelength 525 times 10 to the negative nine and we get three point seven eight times 10 to the negative 19 so let me go ahead and write that down here three point seven eight times 10 to the negative 19 and if you did your units up here you would get joules and so let's think about this number for a second three point seven eight times 10 to the negative 19 is the energy of the photon and that that energy the photon is greater than the work function which means that that is a that's a high-energy photon it's able to knock the electron free because remember this number right here is the minimum amount of energy needed to free the electron and so we've exceeded that minimum amount of energy and so we will produce a photo electron so this this photon is high-energy enough to produce a photo electron so let's go ahead and find the the kinetic energy of the photoelectron that's produced so we're going to use this equation right up here so let me just go and get some more room and I will rewrite that equation so we have the kinetic energy of the photoelectron kinetic energy the photo electron is equal to is equal to the energy of the photon energy the photon - the the work function so let's plug in our numbers the energy the photon was three point seven eight times ten to the negative 19 joules and then the work function is right up here again it's three point four three so minus three point four three times ten to the negative 19 joules so let's get out the calculator again so from that we're going to subtract the work function three point three point four three times ten to the negative 19 and we get three point five times ten to the negative twenty so let's go ahead and write that this is equal to three point five times ten to the negative 20 joules this is equal to the kinetic energy of the photoelectron and we know that kinetic energy is equal to one-half MV squared the problem asked us to solve for the velocity of the photoelectron so all we have to do is plug in the mass of an electron which is nine point one one times 10 to the negative 31st kilograms times V squared this is equal to three point five times ten to the negative twenty so let's do that math so we take three point five times ten to the negative twenty we multiply that by two and then we divide by the mass of an electron nine point one one times ten to the negative 31st and this gives us that that number which we need to take the square root of so square root of our answer gives us the velocity of the electron to point O two points to point eight two times ten to the fifth right so if you look at your decimal place here this would be one two three four five so two point eight times 10 to the fifth meters per second so here's the velocity of the photoelectron produced two point eight times 10 to the fifth meters per second and if you increased the intensity of this light so you had more photons right they would produce more photoelectric so one photon knocks out one photoelectron if it has enough energy to do so so let's uh let's think about the same problem but let's change the wavelength so what if what if your wavelength changed to 625 nanometers so what would happen now well to save time I won't do the calculation but all we would have to do is plug in 625 up here so instead of 525 just plug in 625 to calculate your energy and if you did that so if you use if you use 625 times 10 to the negative nine here I'll go ahead and give you the answer just to save some time you would get 3.2 times 10 to the negative 19 joules and that is lower than the work function all right so let me go ahead and highlight that here so this number this number isn't is not as high as the work function the work function is how much energy we needed to free that electron and since this is lower than the work function that means we do not get a photo electron so you have to have a high enough energy photon in order to produce a photo electron it wouldn't even matter if we increase the intensity right so if we had more and more and more of these photons of this wavelength we still wouldn't produce any photo electrons and so this is the idea of the photoelectric effect which is best explained by thinking about light as a particle

AP® is a registered trademark of the College Board, which has not reviewed this resource.